how many times a number, given by user from keyboard, is occured in the array

-1

The program should calculate: How many times a number, given by user from keyboard, occurs in the array. I have seen similar examples but in this case it is not about the frequency of numbers of the given array, it's about same numbers between users numbers and the given array. So I couldn't make it.

How can we do this just using arrays, if-else condition and for loop?

#include <stdio.h>



int main()

{

    int N=6, size;

    int a  = {1,2,3,4,5,6};    

    int occ [size];

    int i,j;

    int number;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

edited Nov 20 '18 at 7:08

Rai

820418

asked Nov 19 '18 at 23:53

noob

size is not initialized when you use it to declare int occ[size], so how big is the occ array? There are other things wrong but always remember to initialize your variables before you use them. Its easy to just always initialize them (to zero) when you declare them.

– Bwebb
Nov 20 '18 at 0:05

okay i will keep in mind. i know that all code is totally wrong but it was just an idea. i couldn`t find the algorithm for the problem

– noob
Nov 20 '18 at 0:11

if you have the user input you only need to compare that to each element in the array, you should not need the nested for loop. You are currently counting how many times 1 or 2 or 3 or 4 or 5 or 6 are in the array. From the question it sounds like you just to count how many times X is in the array, for some value of X inputted by the user.

– Bwebb
Nov 20 '18 at 0:17

If you compiled with -Wall, it would flag int occ [size]; because at the point where you've placed it, size has yet to be given a value. Move it after the scanf("%d", &size); line. That is, do this: scanf("%d", &size); int occ[size];

– Craig Estey
Nov 20 '18 at 0:20

can you give a sample or expected input-output pair @noob?

– Rai
Nov 20 '18 at 6:29

|
show 1 more comment

-1

How can we do this just using arrays, if-else condition and for loop?

#include <stdio.h>



int main()

{

    int N=6, size;

    int a  = {1,2,3,4,5,6};    

    int occ [size];

    int i,j;

    int number;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

edited Nov 20 '18 at 7:08

Rai

820418

asked Nov 19 '18 at 23:53

noob

size is not initialized when you use it to declare int occ[size], so how big is the occ array? There are other things wrong but always remember to initialize your variables before you use them. Its easy to just always initialize them (to zero) when you declare them.

– Bwebb
Nov 20 '18 at 0:05

okay i will keep in mind. i know that all code is totally wrong but it was just an idea. i couldn`t find the algorithm for the problem

– noob
Nov 20 '18 at 0:11

if you have the user input you only need to compare that to each element in the array, you should not need the nested for loop. You are currently counting how many times 1 or 2 or 3 or 4 or 5 or 6 are in the array. From the question it sounds like you just to count how many times X is in the array, for some value of X inputted by the user.

– Bwebb
Nov 20 '18 at 0:17

If you compiled with -Wall, it would flag int occ [size]; because at the point where you've placed it, size has yet to be given a value. Move it after the scanf("%d", &size); line. That is, do this: scanf("%d", &size); int occ[size];

– Craig Estey
Nov 20 '18 at 0:20

can you give a sample or expected input-output pair @noob?

– Rai
Nov 20 '18 at 6:29

|
show 1 more comment

-1

How can we do this just using arrays, if-else condition and for loop?

#include <stdio.h>



int main()

{

    int N=6, size;

    int a  = {1,2,3,4,5,6};    

    int occ [size];

    int i,j;

    int number;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

edited Nov 20 '18 at 7:08

Rai

820418

asked Nov 19 '18 at 23:53

noob

How can we do this just using arrays, if-else condition and for loop?

#include <stdio.h>



int main()

{

    int N=6, size;

    int a  = {1,2,3,4,5,6};    

    int occ [size];

    int i,j;

    int number;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

c arrays

edited Nov 20 '18 at 7:08

Rai

820418

asked Nov 19 '18 at 23:53

noob

edited Nov 20 '18 at 7:08

Rai

820418

asked Nov 19 '18 at 23:53

noob

edited Nov 20 '18 at 7:08

Rai

820418

edited Nov 20 '18 at 7:08

Rai

820418

edited Nov 20 '18 at 7:08

Rai

820418

asked Nov 19 '18 at 23:53

noob

asked Nov 19 '18 at 23:53

noob

asked Nov 19 '18 at 23:53

noob

size is not initialized when you use it to declare int occ[size], so how big is the occ array? There are other things wrong but always remember to initialize your variables before you use them. Its easy to just always initialize them (to zero) when you declare them.

– Bwebb
Nov 20 '18 at 0:05

okay i will keep in mind. i know that all code is totally wrong but it was just an idea. i couldn`t find the algorithm for the problem

– noob
Nov 20 '18 at 0:11

if you have the user input you only need to compare that to each element in the array, you should not need the nested for loop. You are currently counting how many times 1 or 2 or 3 or 4 or 5 or 6 are in the array. From the question it sounds like you just to count how many times X is in the array, for some value of X inputted by the user.

– Bwebb
Nov 20 '18 at 0:17

If you compiled with -Wall, it would flag int occ [size]; because at the point where you've placed it, size has yet to be given a value. Move it after the scanf("%d", &size); line. That is, do this: scanf("%d", &size); int occ[size];

– Craig Estey
Nov 20 '18 at 0:20

can you give a sample or expected input-output pair @noob?

– Rai
Nov 20 '18 at 6:29

|
show 1 more comment

size is not initialized when you use it to declare int occ[size], so how big is the occ array? There are other things wrong but always remember to initialize your variables before you use them. Its easy to just always initialize them (to zero) when you declare them.

– Bwebb
Nov 20 '18 at 0:05

okay i will keep in mind. i know that all code is totally wrong but it was just an idea. i couldn`t find the algorithm for the problem

– noob
Nov 20 '18 at 0:11

if you have the user input you only need to compare that to each element in the array, you should not need the nested for loop. You are currently counting how many times 1 or 2 or 3 or 4 or 5 or 6 are in the array. From the question it sounds like you just to count how many times X is in the array, for some value of X inputted by the user.

– Bwebb
Nov 20 '18 at 0:17

If you compiled with -Wall, it would flag int occ [size]; because at the point where you've placed it, size has yet to be given a value. Move it after the scanf("%d", &size); line. That is, do this: scanf("%d", &size); int occ[size];

– Craig Estey
Nov 20 '18 at 0:20

can you give a sample or expected input-output pair @noob?

– Rai
Nov 20 '18 at 6:29

size is not initialized when you use it to declare int occ[size], so how big is the occ array? There are other things wrong but always remember to initialize your variables before you use them. Its easy to just always initialize them (to zero) when you declare them.

– Bwebb
Nov 20 '18 at 0:05

okay i will keep in mind. i know that all code is totally wrong but it was just an idea. i couldn`t find the algorithm for the problem

– noob
Nov 20 '18 at 0:11

if you have the user input you only need to compare that to each element in the array, you should not need the nested for loop. You are currently counting how many times 1 or 2 or 3 or 4 or 5 or 6 are in the array. From the question it sounds like you just to count how many times X is in the array, for some value of X inputted by the user.

– Bwebb
Nov 20 '18 at 0:17

If you compiled with -Wall, it would flag int occ [size]; because at the point where you've placed it, size has yet to be given a value. Move it after the scanf("%d", &size); line. That is, do this: scanf("%d", &size); int occ[size];

– Craig Estey
Nov 20 '18 at 0:20

can you give a sample or expected input-output pair @noob?

– Rai
Nov 20 '18 at 6:29

|
show 1 more comment

2 Answers
2

active

oldest

votes

This will solve you problem.

#include <stdio.h>



int main()

{

    int N = 6, size;

    int a  = {1,2,3,4,5,6};    

    int i, j;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    int occ [size];

    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

answered Nov 20 '18 at 7:19

Chameera

628

add a comment |

Your code has some mistakes which have been cleared by @Bwebb so I won't come into that.
Coming back to your problem, this can easily be solved by storing frequency using c++ map container. But since you are trying to implement it using only arrays there is an another way around.

You can make an another array with a fairly large size and initialise all it's elements to 0. Then you take the input from the user for the array and change into it's ASCII value if it's an alphabet when you put it in the index. For everytime the same input comes, you will increase the value of the corresponding index by 1, like this:a[1]++. And whenever you have to find the frequency of a character, you just have to put it in the index of a and print the output.

answered Nov 20 '18 at 0:21

Krab

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

This will solve you problem.

#include <stdio.h>



int main()

{

    int N = 6, size;

    int a  = {1,2,3,4,5,6};    

    int i, j;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    int occ [size];

    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

answered Nov 20 '18 at 7:19

Chameera

628

add a comment |

This will solve you problem.

#include <stdio.h>



int main()

{

    int N = 6, size;

    int a  = {1,2,3,4,5,6};    

    int i, j;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    int occ [size];

    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

answered Nov 20 '18 at 7:19

Chameera

628

add a comment |

This will solve you problem.

#include <stdio.h>



int main()

{

    int N = 6, size;

    int a  = {1,2,3,4,5,6};    

    int i, j;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    int occ [size];

    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

answered Nov 20 '18 at 7:19

Chameera

628

This will solve you problem.

#include <stdio.h>



int main()

{

    int N = 6, size;

    int a  = {1,2,3,4,5,6};    

    int i, j;

    int occured = 0;



    printf ("enter the size of arrayn");

    scanf ("%d", &size);



    int occ [size];

    printf("Enter elements in array:n");

    for (j=0; j<size; ++j){ 

      scanf ("%d", &occ[j]);

    }



    for (j=0; j<size; ++j) {        

      for (i=0; i<N; ++i) {

         if (occ[j] == a[i])

           occured = occured + 1;

      }

    }



    printf ("given numbers occured in array %d times.", occured);

    return 0;

}

answered Nov 20 '18 at 7:19

Chameera

628

answered Nov 20 '18 at 7:19

Chameera

628

answered Nov 20 '18 at 7:19

Chameera

628

answered Nov 20 '18 at 7:19

Chameera

628

add a comment |

answered Nov 20 '18 at 0:21

Krab

add a comment |

answered Nov 20 '18 at 0:21

Krab

add a comment |

answered Nov 20 '18 at 0:21

Krab

answered Nov 20 '18 at 0:21

Krab

answered Nov 20 '18 at 0:21

Krab

answered Nov 20 '18 at 0:21

Krab

answered Nov 20 '18 at 0:21

Krab

add a comment |

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